022 Exam 1 Sample A, Problem 7

Find the slope of the tangent line to the graph of $f(x)=x^{3}-3x^{2}-5x+7$ at the point $(3,-8)$.

ExpandFoundations:
Recall that for a given value, $f'(x)$ is precisely the point of the tangent line through the point $\left(x,f(x)\right)$. Once we have the slope, we can then use the point-slope form for a line:
$y-y_{0}=m(x-x_{0}),$
where $m$ is the known slope and $\left(x_{0},y_{0}\right)$ is a point on the line.

 Solution:

ExpandFinding the slope:
Note that
$f'(x)\,\,=\,\,3x^{2}-6x-5,$
so the tangent line through $(3,-8)$ has slope
$m\,\,=\,\,f'(3)\,\,=\,\,3(3)^{2}-6(3)-5\,\,=\,\,4.$

ExpandWrite the Equation of the Line:
Using the point-slope form listed in foundations, along with the point $(3,-8)$ and the slope $m=4$, we find
$y-(-8)\,\,=\,\,4(x-3),$
or
$y\,\,=\,\,4x-20.$

ExpandFinal Answer:
$y\,\,=\,\,4x-20.$

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